Question

Determine whether the series is convergent or divergent by expressing s

n
​
as a telescoping sum (as in Example 8 ). ∑
n=4
[infinity]
​

n
2
−1
6
​
convergent divergent If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)

Answer 1

The given series is expressed as a telescoping sum: ∑ nⁿ (2n + 1)/6. The series is convergent, and its sum can be given as -2(n(n+1)/3).

Step-by-step explanation:

The given series is ∑ n=4>∞ n2⁄6 - 1

To determine whether this series is convergent or divergent, we need to express it as a telescoping sum.

Let's rewrite each term of the series using a telescoping pattern:

n2⁄6 - 1 = ((n+1)2⁄6) - (n2⁄6) = ((n2 + 2n + 1)⁄6) - (n2⁄6) = (2n + 1)⁄6

Now, we have expressed the series as a telescoping sum: ∑ n=4>∞ (2n + 1)⁄6

Since the terms of this series cancel each other out and the series converges to a finite value when n tends towards infinity, the series is convergent.

To find its sum, we can simplify the expression: (∑ n=4>ⁿ (2n + 1) - 3)(1⁄6)
= (∑ n=4>ⁿ 2n + ∑ n=4>ⁿ 1 - ∑n=4>ⁿ 3)(1⁄6)
= (2(∑ n=4>ⁿ n) + (∑ n=4>ⁿ 1 - (∑ n=4>ⁿ 3)))(1⁄6)
= (2(∑ n=4>ⁿ n) + (∑ n=4>ⁿ 1 - 3(∑ n=4>ⁿ 1)))(1⁄6)

Applying the formulas for the sum of the first n natural numbers, we have:

(2(n(n+1)/2 - 6(1/2n(n+1)/2)))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
= (n(n+1) - 3n(n+1))(1⁄6)
By simplifying the expression further, we obtain:

n(n+1)(1/3 - 1) = n(n+1)(1/3 - 1) = n(n+1)(1/3 - 1) = n(n+1)(-2/3)

Therefore, the sum of the series is -2(n(n+1)/3).